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Advocard [28]
2 years ago
6

Which set of numbers correctly shows all of the factors of 28? 1, 2, 4, 7, 14, 28 1, 2, 7, 14, 28 0, 1, 7, 14, 28 1, 2, 4, 7, 8,

14, 28
Mathematics
2 answers:
Luden [163]2 years ago
8 0

Answer:

1, 2, 4, 7, 14, 28

Step-by-step explanation:

Question:

Which set of numbers correctly shows all of the factors of 28?

1, 2, 4, 7, 14, 28

1, 2, 7, 14, 28

0, 1, 7, 14, 28

1, 2, 4, 7, 8, 14, 28

Solution:

28 = 2^2 * 7

Factors:

1, 2, 2 * 2, 7, 2 * 7, 28

1, 2, 4, 7, 14, 28

Answer: 1, 2, 4, 7, 14, 28

slega [8]2 years ago
3 0

Answer:

1, 2, 4, 7, 14, 28

Step-by-step explanation:

Factors of 28 are the numbers that divide into 28 evenly.

1, 2, 4, 7, 14, 28

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identify the equation that does not belong with the other three. explain your reasoning. 6+x=9 | 15= x+12 | x+9 =11 | 7+x=10
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The function f(x) = 4x + 5 has domain 0sxs 50. What is its range?<br> The range of the function is
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Answer:

  range: 5 ≤ f(x) ≤ 205

Step-by-step explanation:

You can put the extreme values of x in this linear function to see its range:

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2 years ago
I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
masya89 [10]

Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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